Let $G$ be the cartesian product of infinitely many (say countable for simplicity) copies of $S_3$. It is very easy to show that every time you take an element $g$ of prime order in $G$, then there is a complement $H$ to $\langle g\rangle$ in $G$ in the sense that $G=\langle g\rangle H$ and $\langle g\rangle\cap H=1$.

I've read on some russian paper that actually this property holds in all factor groups $G/N$ of $G$, but I'm unable to prove this.

I can reduce easily to the case in which $N\leq G'$ and $\langle gN\rangle\leq G'/N$ but I do not know how to go on from this. Any suggestion?

I consider the property for a group $G$, that every time you take an element $g$ of prime order in $G$, then there is a complement $H$ to $\langle g\rangle$ in $G$ in the sense that $G=\langle g\rangle H$ and $\langle g\rangle\cap H=1$.

Let $G$ be the cartesian product (= unrestricted direct product) of infinitely many (say countable for simplicity) copies of $S_3$. It is very easy to show that $G$ satisfies this property.

I've read on some Russian paper that actually this property holds in all factor groups $G/N$ of $G$, but I'm unable to prove this.

I can reduce easily to the case in which $N\leq G'$ and $\langle gN\rangle\leq G'/N$ but I do not know how to go on from this. Any suggestion?

Let $G$ be the cartesian product of infinitely many (say countable for simplicity) copies of $S_3$. It is very easy to show that every time you take an element $g$ of prime order in $G$, then there is a complement $H$ to $\langle g\rangle$ in $G$ in the sense that $G=\langle g\rangle H$ and $\langle g\rangle\cap H=1$.

I've read on some russian paper that actually this property holds for all factor groups $G/N$ of $G$, but I'm unable to prove this.

I can reduce easily to the case in which $N\leq G'$ and $\langle gN\rangle\leq G'/N$ but I do not know how to go on from this. Any suggestion?

Let $G$ be the cartesian product of infinitely many (say countable for simplicity) copies of $S_3$. It is very easy to show that every time you take an element $g$ of prime order in $G$, then there is a complement $H$ to $\langle g\rangle$ in $G$ in the sense that $G=\langle g\rangle H$ and $\langle g\rangle\cap H=1$.

I've read on some russian paper that actually this property holds in all factor groups $G/N$ of $G$, but I'm unable to prove this.

I can reduce easily to the case in which $N\leq G'$ and $\langle gN\rangle\leq G'/N$ but I do not know how to go on from this. Any suggestion?